Simplify and expand the following expression: $ \dfrac{1}{y - 3}- \dfrac{5}{2y - 18}- \dfrac{3}{y^2 - 12y + 27} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $2$ out of denominator in the second term: $ \dfrac{5}{2y - 18} = \dfrac{5}{2(y - 9)}$ We can factor the quadratic in the third term: $ \dfrac{3}{y^2 - 12y + 27} = \dfrac{3}{(y - 3)(y - 9)}$ Now we have: $ \dfrac{1}{y - 3}- \dfrac{5}{2(y - 9)}- \dfrac{3}{(y - 3)(y - 9)} $ The least common multiple of the denominators is: $ (y - 3)(y - 9)$ In order to get the first term over $(y - 3)(y - 9)$ , multiply by $\dfrac{2(y - 9)}{2(y - 9)}$ $ \dfrac{1}{y - 3} \times \dfrac{2(y - 9)}{2(y - 9)} = \dfrac{2(y - 9)}{(y - 3)(y - 9)} $ In order to get the second term over $(y - 3)(y - 9)$ , multiply by $\dfrac{y - 3}{y - 3}$ $ \dfrac{5}{2(y - 9)} \times \dfrac{y - 3}{y - 3} = \dfrac{5(y - 3)}{(y - 3)(y - 9)} $ In order to get the third term over $(y - 3)(y - 9)$ , multiply by $\dfrac{2}{2}$ $ \dfrac{3}{(y - 3)(y - 9)} \times \dfrac{2}{2} = \dfrac{6}{(y - 3)(y - 9)} $ Now we have: $ \dfrac{2(y - 9)}{(y - 3)(y - 9)} - \dfrac{5(y - 3)}{(y - 3)(y - 9)} - \dfrac{6}{(y - 3)(y - 9)} $ $ = \dfrac{ 2(y - 9) - 5(y - 3) - 6} {(y - 3)(y - 9)} $ Expand: $ = \dfrac{2y - 18 - 5y + 15 - 6}{2y^2 - 24y + 54} $ $ = \dfrac{-3y - 9}{2y^2 - 24y + 54}$